Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following:
- 1.p + q + r − s − t is definitely even.
- 2.2p + q + 2r − 2s + t is definitely odd.
Which of the above statements is/are correct?
Answer & explanation
Answer: (a) 1 only
Subtracting a number changes odd/even exactly as adding it does, so p + q + r − s − t has the same parity as the sum of all five, which is even (it contains two odd numbers). In 2p + q + 2r − 2s + t the doubled terms are even, leaving the parity of q + t, which may be even.
- Adding or subtracting a number has the same effect on parity, so signs can be ignored.
- Statement 1: parity of p + q + r − s − t = parity of p + q + r + s + t. Two odds make an even and evens stay even, so the result is always even. Correct.
- Statement 2: 2p, 2r and 2s are even, so the parity equals that of q + t.
- Example: p = 1 and s = 1 are the odd ones, q = r = t = 2. Then 2 + 2 + 4 − 2 + 2 = 8, which is even. So statement 2 is not always true.
- ✓ 1. Its parity is that of the sum of all five numbers: three evens and two odds add up to an even number, whatever the order.
- ✗ 2. Only q and t decide the parity; if both are even (or both odd) the expression is even, so it is not definitely odd.
Remember · For parity, treat minus as plus and ignore every term multiplied by an even number.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·